2779. Maximum Beauty of an Array After Applying Operation - Detailed Explanation

Problem Statement

You are given a 0-indexed integer array nums and a non-negative integer k. In one operation, you can select an index i (that hasn't been chosen before) and replace nums[i] with any integer in the range [nums[i] - k, nums[i] + k]. In other words, you can adjust one array element up or down by at most k. You may perform this operation on as many different indices as you like (including zero operations or up to all indices, but each index at most once).

The beauty of an array is defined as the length of the longest subsequence in which all elements are equal. (A subsequence is any sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.)

The task is to return the maximum possible beauty of array nums after applying the allowed operation any number of times (within the given constraints). In simpler terms, determine the largest number of elements that can be made equal by appropriately adjusting some of the elements by at most k each.

Constraints

• 1 <= nums.length <= 10^5 – The array can be quite large (up to 100,000 elements).
• 0 <= nums[i] <= 10^5 and 0 <= k <= 10^5 – Array values and k are non-negative and can be as large as 100,000.
• You can apply the operation on each index at most once. (If k = 0, the operation doesn’t change any value.)

These constraints imply that an efficient solution is needed (around O(n log n) or O(n) time), because a brute-force approach (e.g., checking all subsequences) would be infeasible for n up to 100k.

Example Test Cases

Example 1:

• Input: nums = [4, 6, 1, 2], k = 2
• Output: 3
• Explanation: We can perform operations to maximize the number of equal elements. For instance, choose index 1 (value 6) and replace nums[1] with 4 (within range [4, 8]), and choose index 3 (value 2) and replace nums[3] with 4 (within range [0, 4]). After these operations, nums becomes [4, 4, 1, 4]. The longest subsequence of equal values is three 4's (indices 0, 1, and 3). Thus, the beauty of the array is 3. It can be proven that 3 is the maximum achievable length.

Example 2:

• Input: nums = [1, 1, 1, 1], k = 10
• Output: 4
• Explanation: All elements are already equal (all are 1), so the beauty is 4 (the whole array) without any operations. Even though k is large, we don’t need to change anything.

Example 3:

• Input: nums = [3, 1, 3, 5], k = 0
• Output: 2
• Explanation: With k = 0, no element can be changed at all. The array stays the same, so we can only rely on existing equal values. The value 3 appears twice, which is the longest equal-value subsequence (there are two 3’s). Thus the maximum beauty is 2.

Example 4:

• Input: nums = [2, 4, 7, 1], k = 3
• Output: 4
• Explanation: Here nums can be fully equalized. We can adjust each number to 4 (for example: 2 → 4, 4 → 4, 7 → 4, 1 → 4, since 2,7 can move to 4 within ±3 and 1 can move to 4 within +3). After operations, all elements become 4, so the beauty is 4 (the entire array).

These examples illustrate different scenarios: doing no operations, adjusting some elements, having no ability to adjust, or even making the entire array equal. Our goal is to find a general method to compute the maximum beauty for any input.

Understanding the Problem

The core challenge is to find the largest group of array elements that can be made equal through the allowed operations. Because each element can be adjusted by at most k up or down, two elements can potentially meet at the same value if their original values differ by no more than 2*k. For example, if one element is 6 and another is 4 with k=2, 6 can be lowered to 4 and 4 can stay the same (difference 2 ≤ 2*k), so they can both become 4. If the difference is larger than 2*k, then even at maximum adjustments one high and one low value cannot meet at a common value.

Another way to view the problem is: what is the largest subset of nums such that the difference between the maximum and minimum element in that subset is at most 2*k? Any such subset’s elements can all be adjusted to the value of the smallest element boosted by k, or the largest element reduced by k, or some value in between, achieving equality. If a subset has a larger spread than 2*k, it’s impossible to make all those elements equal with the allowed ±k adjustments.

So effectively, we are looking for the longest subsequence (not necessarily contiguous in the original array) where the difference between the min and max elements is ≤ 2*k. The length of that subsequence is the maximum beauty we can achieve.

Hints

If you're struggling to come up with a solution, consider these hints to guide your thinking:

1. Think about sorting: It might be easier to compare elements and find groups that can be equalized if the array is sorted first. Sorting the array can help cluster together values that are close to each other.

2. Use the two-pointer or sliding window technique: After sorting, try to find the longest contiguous segment (window) in the sorted array where the difference between the largest and smallest element in that segment is at most 2*k. This segment corresponds to a set of elements that can all be made equal.

3. Why 2*k? Remember that each element can move up or down by at most k. If you have a segment from index i to j (in sorted order), the smallest value is nums[i] and the largest is nums[j]. All values in between are within this range as well. For all these to meet, you need nums[j] - nums[i] ≤ 2*k. If this holds, the entire group can converge to some value in between. If it doesn't hold, the window is too wide and you should adjust by moving one end of the window.

4. Try simple cases: If k = 0, the problem reduces to finding the frequency of the most common element (since no changes can be made). If k is extremely large (e.g., larger than half the range of values in nums), the answer might simply be the length of the array (because you can make everything equal). These extreme cases can guide your understanding for the general case.

Using these hints, see if you can derive an approach before reading on.

Approach 1: Brute Force (Conceptual Understanding)

A brute force approach to this problem is mainly useful for understanding the problem, because it would be too slow for the given constraints. Here are a couple of brute force ideas and why they’re not feasible in practice:

• Brute Force by Checking Subsequences: You could, in theory, generate all subsequences of the array and check if a given subsequence can be made all equal with the allowed operations. Then pick the longest valid one. However, there are an exponential number of subsequences (2^n possible subsequences for n elements), so this approach is completely impractical even for moderate n.

• Brute Force by Target Values: Another idea: consider possible target values that the array elements could all be turned into. For each possible target value X, count how many elements of nums can be adjusted to exactly X (meaning for each num in nums, check if X lies in [num - k, num + k]). The target that yields the highest count would give the maximum beauty. The problem is that X could be any integer in a certain range. In the worst case, nums[i] and k can be up to 100,000, so X could range roughly from min(nums) - k to max(nums) + k (potentially up to ~200,000 range or more). Checking every possible target in that range (which could be 200k possibilities) for up to 100k elements each would be on the order of 20 billion checks – far too many to compute efficiently.

• Brute Force by Pairwise Comparison (Nested Loops): A simpler brute force is to sort the array and then use two nested loops to find the longest segment where the difference condition holds. For each starting index i, you expand an ending index j as far as possible such that nums[j] - nums[i] ≤ 2*k. This is essentially what the two-pointer method (next approach) does, but doing it with two nested loops naively would result in O(n^2) time in the worst case (e.g., if k is large enough that the inner loop almost always runs to the end for each i). With n up to 100k, n^2 (10^10 operations) is too slow.

Why we mention this approach: Brute force helps confirm our understanding: we’re trying to group numbers such that their max-min difference is within 2*k. It also highlights the need for a more efficient strategy. We need to avoid examining every pair or every possible target explicitly.

Approach 2: Sorting and Two-Pointer Sliding Window (Optimal Solution)

The optimal solution uses a sorting step and a two-pointer (sliding window) technique to efficiently find the largest group of values that satisfy the condition. This approach is both intuitive and runs fast enough for the problem constraints.

Idea:

1. Sort the array nums. Sorting the array helps because any potential group of equalized numbers will appear as a contiguous segment in sorted order. If you take any set of numbers that can all be made equal, their sorted order will cluster them together (since extremely large or small values that can’t join the group will be outside the range). After sorting, if we look at a window nums[i...j] (from index i to j in the sorted array), this represents a candidate subsequence (not necessarily contiguous in the original array, but that doesn’t matter for a subsequence). We want to ensure in that window the difference nums[j] - nums[i] is not more than 2*k.

2. Use two pointers (or indices) to maintain a sliding window of valid candidates:

   • Have one pointer (or index) l for the start of the window and another r for the end of the window. Initially, both can start at 0 (the beginning of the sorted array).

   • We will expand r step by step (r moves to the right) to consider including more elements in the window. For each new position of r, check if the window [l, r] is still valid under the condition. The key condition is:
     Is nums[r] - nums[l] ≤ 2*k?
     If this condition holds, it means all elements from index l to r in the sorted array have a max-min difference ≤ 2*k, so they can all be made equal by some appropriate adjustments.

   • If adding the new element at r violates the condition (meaning nums[r] - nums[l] > 2*k), then the window has become invalid (too wide a range). To fix this, we need to shrink the window from the left by incrementing l (move the left pointer to the right). We keep moving l up until the window is valid again. Because the array is sorted, moving l to l+1 will reduce the range of the window (the minimum value in the window increases), helping restore validity. Essentially, we are sliding the window along the array.

   • Throughout this process, we track the maximum window size (the number of elements in the window) that satisfies the condition. That maximum size is our answer (maximum beauty).

3. Why this works: When the array is sorted, any window that satisfies nums[r] - nums[l] ≤ 2*k represents a set of values that can all be adjusted to a common number. If the window is as large as possible, that’s the largest equalizable subsequence. The two-pointer technique efficiently finds the largest such window by never moving backwards – each of l and r only moves forward through the array at most once, making the algorithm linear in time once the array is sorted.

Step-by-Step Process:

• Sort nums in non-decreasing order.
• Initialize two pointers: l = 0 and r = 0. Also initialize max_length = 0 (to keep track of the longest valid window found).
• Loop r from 0 to n-1 (inclusive):

  • For each r, ensure the window [l, r] is valid. While the window is invalid (i.e., nums[r] - nums[l] > 2*k), increment l (move the left boundary to the right) to shrink the window until it becomes valid again. Because the array is sorted, nums[r] is the maximum in the window and nums[l] is the minimum, so this check ensures the max-min difference condition.

  • Once the window [l, r] satisfies the condition, calculate its length: window_length = r - l + 1. Update max_length = max(max_length, window_length).

• After the loop, max_length will hold the length of the largest window found that satisfies the difference condition, which is the answer.

This approach is effectively a sliding window approach on a sorted array, where we maintain the largest window with max - min <= 2*k. It runs in O(n) time for the sliding window after sorting, and sorting takes O(n log n), so overall complexity is O(n log n). This is efficient for n up to 100k.

Let's apply this method to a concrete example to see how it works step by step.

Step-by-Step Example Walkthrough

Take Example 1: nums = [4, 6, 1, 2], k = 2. We will walk through the sliding window approach:

1. Sort the array: Sorted nums becomes [1, 2, 4, 6].
2. Initialize: l = 0, max_length = 0. We will iterate r from 0 to 3 (indices of sorted array).

• Iteration r = 0:

  • Window = [l, r] = [0, 0] which is just the element [1].
  • Check validity: nums[r] - nums[l] = nums[0] - nums[0] = 1 - 1 = 0, which is ≤ 2*k = 4. Valid.
  • Window length = 0 - 0 + 1 = 1. Update max_length = max(0, 1) = 1.
    (Window [1] is valid – of course, any single element is trivially equalizable with itself.)

• Iteration r = 1:

  • Window = [0, 1] corresponding to elements [1, 2].
  • Check: nums[1] - nums[0] = 2 - 1 = 1 ≤ 4. Still valid.
  • Window length = 1 - 0 + 1 = 2. Update max_length = max(1, 2) = 2.
    (Elements 1 and 2 can be made equal because their difference is 1 which is ≤ 4; e.g., 1 could become 2 or 2 could become 1 or both meet at some 1.x value within range – the exact meeting value isn’t needed, just knowing they can meet is enough.)

• Iteration r = 2:

  • Window = [0, 2] -> elements [1, 2, 4].
  • Check: nums[2] - nums[0] = 4 - 1 = 3 ≤ 4. Valid.
  • Window length = 2 - 0 + 1 = 3. Update max_length = max(2, 3) = 3.
    (Elements 1, 2, 4 can all be made equal? Yes, because max=4, min=1, difference 3 ≤ 4. For instance, 1 can become 3, 2 can become 3, 4 can become 3 (all within ±2). So 1,2,4 can meet at value 3.)

• Iteration r = 3:

  • Window = [0, 3] -> elements [1, 2, 4, 6].

  • Check: nums[3] - nums[0] = 6 - 1 = 5, which is > 4. This window is not valid because the range (5) exceeds 2*k. The element at r (which is 6) is too far apart from the element at l (which is 1) to all meet within ±2 adjustments.

  • To fix this, move l forward: set l = 1. Now the window is [1, 3] -> elements [2, 4, 6].

    • Re-check validity with the new window: nums[3] - nums[1] = 6 - 2 = 4 ≤ 4. Now it’s valid. We had to drop the previous l (value 1) from the window.

  • Now with l = 1 and r = 3, window length = 3 - 1 + 1 = 3. max_length is currently 3, so it remains 3.
    (So the largest valid window ends up being of length 3. In sorted terms that’s [2,4,6], and indeed in the original array those correspond to values {2,4,6} which can all become 4 as we saw. The element 1 was the outlier that had to be excluded to allow 6 to join the party.)

• End of loop: max_length = 3. This matches our expectation and the output for the example.

Throughout the sliding window process, each element was considered and the window adjusted as needed. The largest window we found was of size 3, meaning 3 elements can be made equal simultaneously. In the example, that corresponds to the value 4 appearing three times after operations, which indeed was the solution we found manually.

This process works efficiently for large arrays because each element enters and leaves the window at most once. The two-pointer method ensures we don’t do redundant comparisons.

Complexity Analysis

• Brute Force Approach: The conceptual brute force methods would be extremely slow:

  • Checking all subsequences is O(2^n) which is infeasible for n > 20 or so.
  • Trying all possible target values would be on the order of O(n * R) where R is the range of possible values to check (potentially hundreds of thousands), leading to ≈10^10 operations in worst case (n=100k, R ~ 200k). This is also infeasible.
  • A double loop scanning for valid windows is O(n^2) in worst case (if the window is usually valid until the end, the inner loop runs ~n for each of n iterations of the outer loop). For n = 100k, n^2 = 10^10, again too slow.
  • Space complexity for these is O(1) (no extra space), but time is the limiting factor.

• Sorting + Two Pointers Approach: Sorting the array takes O(n log n) time. The two-pointer sliding window scan is O(n) time since each pointer l and r moves at most n steps total. Therefore, the overall time complexity is O(n log n + n) = O(n log n). For n = 100k, this is very manageable (on the order of a million operations or a few million, depending on constants). The space complexity is O(1) extra (if sorting in place) or O(n) if the sorting algorithm needs extra buffer or if we create a copy. This is fine for n = 100k.

In summary:

• Brute Force: ~O(n^2) time (or worse), O(1) space.
• Optimal (Sorting + Two Pointers): ~O(n log n) time, O(n) space (or O(1) extra).

Only the optimal approach is feasible for the given constraints, but understanding the brute force reasoning helps validate the strategy.

Code Implementation (Python)